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aivan3 [116]
2 years ago
5

Part 2: a) Write VHDL code for a top module that invokes necessary components to display the four decimal digits on four seven-s

egment displays of a NEXYX 4 FPGA. The top module takes the 100 MHz FPGA clock, a reset signal, an enabler signal as the inputs. b) Makes necessary changes in a constraint file to implement your design on the FPGA and present only the uncommented parts of the constraint file in your exam script.
Computers and Technology
1 answer:
dimulka [17.4K]2 years ago
7 0
I don’t know ‍♀️ sorry
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It takes 2 seconds to read or write one block from/to disk and it also takes 1 second of CPU time to merge one block of records.
Alexxx [7]

Answer:

Part a: For optimal 4-way merging, initiate with one dummy run of size 0 and merge this with the 3 smallest runs. Than merge the result to the remaining 3 runs to get a merged run of length 6000 records.

Part b: The optimal 4-way  merging takes about 249 seconds.

Explanation:

The complete question is missing while searching for the question online, a similar question is found which is solved as below:

Part a

<em>For optimal 4-way merging, we need one dummy run with size 0.</em>

  1. Merge 4 runs with size 0, 500, 800, and 1000 to produce a run with a run length of 2300. The new run length is calculated as follows L_{mrg}=L_0+L_1+L_2+L_3=0+500+800+1000=2300
  2. Merge the run as made in step 1 with the remaining 3 runs bearing length 1000, 1200, 1500. The merged run length is 6000 and is calculated as follows

       L_{merged}=L_{mrg}+L_4+L_5+L_6=2300+1000+1200+1500=6000

<em>The resulting run has length 6000 records</em>.

Part b

<u><em>For step 1</em></u>

Input Output Time

Input Output Time is given as

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}

Here

  • L_run is 2300 for step 01
  • Size_block is 100 as given
  • Time_{I/O per block} is 2 sec

So

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}\\T_{I.O}=\frac{2300}{100} \times 2 sec\\T_{I.O}=46 sec

So the input/output time is 46 seconds for step 01.

CPU  Time

CPU Time is given as

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}

Here

  • L_run is 2300 for step 01
  • Size_block is 100 as given
  • Time_{CPU per block} is 1 sec

So

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}\\T_{CPU}=\frac{2300}{100} \times 1 sec\\T_{CPU}=23 sec

So the CPU  time is 23 seconds for step 01.

Total time in step 01

T_{step-01}=T_{I.O}+T_{CPU}\\T_{step-01}=46+23\\T_{step-01}=69 sec\\

Total time in step 01 is 69 seconds.

<u><em>For step 2</em></u>

Input Output Time

Input Output Time is given as

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}

Here

  • L_run is 6000 for step 02
  • Size_block is 100 as given
  • Time_{I/O per block} is 2 sec

So

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}\\T_{I.O}=\frac{6000}{100} \times 2 sec\\T_{I.O}=120 sec

So the input/output time is 120 seconds for step 02.

CPU  Time

CPU Time is given as

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}

Here

  • L_run is 6000 for step 02
  • Size_block is 100 as given
  • Time_{CPU per block} is 1 sec

So

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}\\T_{CPU}=\frac{6000}{100} \times 1 sec\\T_{CPU}=60 sec

So the CPU  time is 60 seconds for step 02.

Total time in step 02

T_{step-02}=T_{I.O}+T_{CPU}\\T_{step-02}=120+60\\T_{step-02}=180 sec\\

Total time in step 02 is 180 seconds

Merging Time (Total)

<em>Now  the total time for merging is given as </em>

T_{merge}=T_{step-01}+T_{step-02}\\T_{merge}=69+180\\T_{merge}=249 sec\\

Total time in merging is 249 seconds seconds

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3 years ago
Which of the following best describes the purpose of child labor laws?
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Answer:

Are there supposed to be answer choices with this question?

Child labor laws are basically laws placing restrictions and regulations on the work of minors.

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Which of these port transmits digital video
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Answer:

uhm

Explanation:

theres nothing there

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When is it not necessary to invoke the method myMethod using dot notation? When myMethod is private. When myMethod is invoked by
Paraphin [41]

Answer:

When myMethod is invoked by a method in the same class as myMethod.

Explanation:

When you are in a class, it is like a different enviroment and within that class you can call a method without using the dot notation because you are still within the class.

Outside the class and despite the type of method, public, private or static, the dot notation will be required.

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Question A low level distributed denial of service (DDoS) attack that involves SYN or SYN/ACK flooding describes what type of at
Morgarella [4.7K]

Answer:

TCP SYN flood

Explanation:

TCP SYN flood (a.k.a. SYN flood) is a type of Distributed Denial of Service (DDoS) attack that exploits part of the normal TCP three-way handshake to consume resources on the targeted server and render it unresponsive.

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2 years ago
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